(-x + 3) + (x + 5) = i need a step by step explanation plz

Compute the directional derivative of the following function at the given point P in the direction of the given vector. Be sure

borishaifa [10]

Answer:

The directional derivate is given by: $D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$

Step-by-step explanation:

The directional derivative at point (x,y) is given by:

$D_{u}(x,y) = f_{x}(x,y)*a + f_{y}(x,y)*b$

In which a is the x component of the unit vector and b is the y component of the unit vector.

Vector:

We are given the following vector: $v = (3,1)$

Its modulus is given by: $\sqrt{3^2 + 1^2} = \sqrt{10}$

The unit vector is given by each component divided by it's modulus. So

$v_u = (\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}})$

This means that $a = \frac{3}{\sqrt{10}}, b = \frac{1}{\sqrt{10}}$

Partial derivatives:

$f(x,y) = \ln{(2 + 3x^2 + 3y^2)}$

So

$f_x(x,y) = \frac{6x}{\ln{(2 + 3x^2 + 3y^2)}}$

$f_x(1,-2) = \frac{6(1)}{\ln{(2 + 3(1)^2 + 3(-2)^2)}} = \frac{6}{\ln{17}}$

$f_y(x,y) = \frac{6y}{\ln{(2 + 3x^2 + 3y^2)}}$

$f_y(1,-2) = \frac{6(-2)}{\ln{(2 + 3(1)^2 + 3(-2)^2)}} = -\frac{12}{\ln{17}}$

Directional derivative:

$D_{u}(x,y) = f_{x}(x,y)*a + f_{y}(x,y)*b$

$D_{u}(x,y) = \frac{6}{\ln{17}}\times\frac{3}{\sqrt{10}}-\frac{12}{\ln{17}}\times\frac{1}{\sqrt{10}}$

$D_{u}(x,y) = \frac{18}{\ln{17}\sqrt{10}} - \frac{12}{\ln{17}\sqrt{10}}[tex][tex]D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$

The directional derivate is given by: $D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$

8 0

3 years ago

What is (5y)^3 without exponents

Kruka [31]

Answer:

125y^3

Step-by-step explanation:

5^3= 5*5*5=125

y^3

5 0

3 years ago

Read 2 more answers

The company you work for has a policy of paying overtime for all hours worked in excess of 40 per week you worked 8 hours each d

makvit [3.9K]

you would have earned 2 hours worth

5 0

3 years ago

A uniform beam of length L = 7.30m and weight = 4.45x10²N is carried by two ovorkers , Sam and Joe - Determine the force exert e

Mama L [17]

Answer:

Effort and distance = Load x distance

7.30 x 4.45x10^2N = 3.2485 X 10^3N

We then know we can move 3 points to the right and show in regular notion.

= 3248.5

Divide by 2 = 3248.5/2 = 1624.25 force

Step-by-step explanation:

In the case of a Second Class Lever as attached diagram shows proof to formula below.

Load x distance d1 = Effort x distance (d1 + d2)

The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :

Effort x its distance from wheel centre = Load x its distance from the wheel centre.

This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)

This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :

Therefore F x d = R2 x D

It can be seen that this is so if we imagine taking away the Reaction R2.

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.

The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.

We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.

So if CM = ACM then F x d = R2 x D

from our kitchen scales example above 2kg x 0.5m = R2 x 1m

so R2 = 1m divided by 2kg x 0.5m

therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.

Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.

In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m

CM = ACM

(F1 x d1) + (F2 x d2) = R2 x Length of beam

(60kg x 6m) + (40Kg x 9m) = R2 x 12m

(60kg x 6m) + (40Kg x 9m) / 12m = R2

360kgm + 360km / 12m = R2

720kgm / 12m = R2

60kg = R2 (note the unit 'm' for metres is cancelled out)

So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg

4 0

3 years ago

If you get this, you are a critical thinker.

vazorg [7]

Answer:

The correct answer is 34 people

Step-by-step explanation:

Killing 3 people does not change the number of people in the bedroom, it only means that there are 30 dead people and 4 living people, but overall, here are 34 people in the room.

4 0

3 years ago